# RD Sharma Solutions Class 8 Practical Geometry Exercise 18.4

## RD Sharma Solutions Class 8 Chapter 18 Exercise 18.4

#### Exercise 18.4

1. Construct a quadrilateral ABCD, in which AB = 6 cm, BC = 4 cm, CD = 4 cm, $\angle B=95°\;$ and $\angle C=90°\;$.

Steps of construction:

Step I: Draw BC = 4 cm.

Step II: Construct $\angle ABC=95°\;$ at B.

Step III: With B as the center and radius 6 cm, cut off BA = 6 cm.

Step IV: Construct $\angle BCD=90°\;$ at C.

Step V: With C as the center and radius 4 cm, cut off BA = 4 cm.

Step VI: Join CD.

2. Construct a quadrilateral ABCD, where AB = 4.2 cm, BC = 3.6 cm, CD = 4.8 cm, $\angle B=30°\;$ and $\angle C=150°\;$.

Steps of construction:

Step I: Draw BC = 3.6 cm.

Step II: Construct $\angle ABC=30°\;$ at B.

Step III: With B as the center and radius 4.2 cm, cut off BA = 4.2 cm.

Step IV: Construct $\angle BCD=150°\;$ at C.

Step V: With C as the center and radius 4.8 cm, cut off CD = 4.8 cm.

3. Construct a quadrilateral PQRS, in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, $\angle Q=75°\;$ and $\angle R=120°\;$.

Steps of construction:

Step I: Draw QR = 2.5 cm.

Step II: Construct $\angle PQr=75°\;$ at Q.

Step III: With Q as the center and radius 3.5 cm, cut off QP = 3.5 cm.

Step IV: Construct $\angle QRS=120°\;$ at R.

Step V: With R as the center and radius 4.1 cm, cut off RS = 4.1 cm.

Step VI: Join PS.

4. Construct a quadrilateral ABCD given BC = 6.6 cm, CD = 4.4 cm, AD = 5.6 cm and $\angle D=100°\;$ and $\angle C=95°\;$.

Steps of construction:

Step I: Draw DC = 4.4 cm.

Step II: Construct $\angle ADC=100°\;$ at D.

Step III: With D as the center and radius 5.6 cm, cut off DA = 5.6 cm.

Step IV: Construct $\angle BCD=95°\;$ at C.

Step V: With C as the center and radius 6.6 cm, cut off CB = 6.6 cm.

Step VI: Join AB.

5. Construct a quadrilateral ABCD, in which AD = 3.5 cm, AB = 4.4 cm, BC = 4.7 cm, $\angle A=125°\;$ and $\angle B=120°\;$.

Steps of construction:

Step I: Draw AB = 4.4 cm.

Step II: Construct $\angle BAD=125°\;$ at A.

Step III: With A as the centre and radius 3.5 cm, cut off AD = 3.5 cm.

Step IV: Construct $\angle ABC=125°\;$ at B.

Step V: With B as the centre and radius 4.7 cm, cut off BC = 4.7 cm.

Step VI: Join CD.

6. Construct a quadrilateral PQRS, in which $\angle Q=45°\;$, $\angle R=70°\;$, QR = 5 cm, PQ = 9 cm and RS = 7 cm.

Steps of construction:

Step I: Draw QR = 5 cm.

Step II: Construct $\angle PQR=45°\;$ at Q.

Step III: With Q as the center and radius 9 cm, cut off QP = 9 cm.

Step IV: Construct $\angle QRS=90°\;$ at R.

Step V: With R as the center and radius 7 cm, cut off RS = 7 cm.

Since, the line segment PQ and RS intersect each other, the quadrilateral cannot be constructed.

7. Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, $\angle A=90°\;$ and $\angle B=105°\;$.

Steps of construction:

Step I: Draw AB = 3 cm.

Step II: Construct $\angle DAB=90°\;$ at A.

Step III: With A as the center and radius 5 cm, cut off AD = 5 cm.

Step IV: Construct $\angle ABC=105°\;$ at B.

Step V: With B as the center and radius 3 cm, cut off BC = 3 cm.

Step VI: Join CD.

8. Construct a quadrilateral BDEF, where DE = 4.5 cm, EF = 3.5 cm, FB = 6.5 cm, $\angle F=50°\;$ and $\angle E=100°\;$.

Steps of construction:

Step I: Draw EF = 3.5 cm.

Step II: Construct $\angle DEF=100°\;$ at E.

Step III: With E as the center and radius 4.5 cm, cut off DE = 4.5 cm.

Step IV: Construct $\angle EFB=50°\;$ at F.

Step V: With F as the center and radius 6.5 cm, cut off FB = 6.5 cm.

Step VI: Join BD.